How do i find the equation of line that is perpendicular on line 4x+y-1=0 and goes through the intersection from line 2x-5y+3=0 and x-3y-7=0 ?

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Can you explain in detail how I get there ?

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Answer 1 · 2017-10-03T17:14:37

#x-4y-24=0# is the required equation of the straight line.

The equation of the line perpendicular to #ax+by+c=0# is #bx-ay+lamda=0#

Remember
#rarr# Constant is different.

#rarr# Coefficients are exchanged.

#rarr# Sign between x and y is changed.

The equation of line perpendicular to #4x+y-1=0# is # x-4y+lamda=0#

The point of intersection of straight lines #2x-5y+3=0# and #x-3y-7=0# is #(-44,-17)#

Putting #x=-44# and #y=-17# in #x-4y+lamda=0# we get,

Or, #-44-4*(-17)+lamda=0#

Or, #lamda+68-44=0#

Or, #lamda=-24#

So the required equation of the straight line is #x-4y-24=0#.